Motion in One Dimension

Transcription

1 Motion in One Dimension Intoduction: In this lab, you will investigate the motion of a olling cat as it tavels in a staight line. Although this setup may seem ovesimplified, you will soon see that a detailed undestanding of linea motion is at the heat of many impotant concepts in physics. In this lab, you will also become familia with the LabPo inteface and techniques fo gaphing position, velocity, and acceleation in eal time. Finally, you will pactice using diffeentiation and integation to investigate the elationships between position, velocity, and acceleation. Mateials Blue cat Set of weights 2-mete sticks affixed to table to fom tack Low-fiction pulley LabPo inteface with motion detecto Blue painte s tape Setup and pocedue You cat and tack should be set up and eady to go. The set up should appea simila to the illustation above. All you need to do befoe stating is get the compute and LabPo inteface eady to take data. Eveything you need is in the plastic LabPo box. 1. Pull out the LabPo inteface. It is teal and about the size of a gaphing calculato. Plug the AC adapto into the LabPo and then into the wall socket. 1 of 15

2 2. Locate the motion detecto. Use a cable that has two white ends to connect the motion detecto to a DIGI pot on the LabPo. Place the motion detecto upight on the table so that the speake is pointing staight down the tack. 3. Use a USB cable to connect the LabPo to the compute. 4. Open the Expeiments folde on the Desktop. 5. Open the file 1D Acceleation.cmbl. You ae now eady to take data. Click on the geen collect button at the uppe ight cone of the sceen. This should make the motion detecto stat to click. If this is not woking, ask fo help fom you instucto. Move the cat aound while collecting to make sue that the motion detecto is woking popely. 6. The motion detecto has a limited ange. If you cat gets too close o too fa, the detecto will not ead the coect distance. Move the cat towad the senso while collecting data until the senso stops eading the coect distance. Mak this spot on the ules using blue tape and neve use any data that is taken when the cat is beyond this point fo the est of the lab. 7. Now move the cat as fa as you can away fom the senso until it stops eading the coect distance. Mak this spot on the ules. Pat I Constant Motion. Pocedue 1. Click on the stopwatch button at the uppe ight cone of the sceen. Set the pogam to collect data fo 60 seconds. 2. Choose one team membe to move the cat. Stat the cat close to the motion detecto and move it slowly away fom the detecto at a constant speed. You should watch the sceen as you do this and obseve the shapes of the gaphs. The position vs. time gaph should be a staight, slanted line, and the velocity vs. time gaph should be a hoizontal line. 3. Once you each the motion senso s limit, stop the cat fo a few seconds. Then bing the cat slowly back towad the senso, again at a constant speed. Repeat fo the duation of the thity seconds. Use File Save As to save you data with a unique file name. 4. Repeat steps two-thee with a diffeent team membe, and move the cat moe quickly than befoe. Again, save you data. Analysis Follow these steps fo both of you data sets: 1. Choose one staight-line segment of the position vs. time gaph. Highlight this egion. 2. Use Analyze Linea Fit to fit a staight line to you data. When you pefom this step, the compute is finding the equation of line with slope and intecept that best match you data. 3. Wite out the equation fo position vs. time fo constant-velocity motion. Explain why the slope of the linea fit coesponds to velocity. 2 of 15

3 4. Highlight the coesponding egion of the velocity vs. time gaph. 5. Use Analyze Statistics to find the mean of the velocity. 6. Do you values fo velocity fom steps 2 and 5 match? 7. Highlight the coesponding egion of the acceleation vs. time gaph. Use Analyze Statistics to find the mean of the acceleation. 8. Explain what this value of acceleation means. Once you have completed the analysis, pint up each gaph. Also answe the following: 9. How did the velocity fo the second un compae with that of the fist? 10. Did the faste motion coespond to a steepe o shallowe slope on the position vs. time gaph? Pat II Acceleated Motion Speeding Up. Investigating Deivatives. Pocedue 1. Use the stopwatch button at the uppe ight cone of the sceen to change the data collection duation to 8 seconds. 2. Pull the thead that is attached to the cat though the pulley at the end of the table. Place a 500g weight on the cat, and get a 50g weight eady to hang fom the loop at the othe end of the thead. Place the cat next to the motion senso. 3. Have one team membe hit collect. Have anothe team membe place the 50g weight in the loop and elease the cat. Repeat until you get a clean gaph. Analysis 1. Highlight the good egion of data on you gaph. You do not want to pay any attention to data that was collected befoe the cat was eleased o afte the weight hit the gound. 2. Use Analyze Cuve Fit to have the pogam find the paabola that best matches you data. Choose Quadatic, then click, Ty Fit. If the cuve seems to match you data well, click OK, othewise ask you instucto fo assistance. 3. Wite out the equation fo position as a function of time fo acceleated motion. Identify how each pat of the cuve fit elates to this equation, and fill in numbes in the table below. x(t) = a v 0 x 0 3 of 15

4 4. Woking by hand, take the deivative of both sides of the equation above. This should give you an equation fo velocity as a function of time. This equation should match the shape of you velocity vs. time gaph. v(t) = dx/dt = 5. Use Analyze Linea Fit to fit a staight line to you velocity vs. time gaph. Detemine how each pat of the linea fit elates to the equation above, and fill in the numbes in the following table. a v 0 6. Woking by hand take the deivative of the equation above. This will give you an equation fo acceleation as a function of time. You equation should be vey, vey simple. a(t) = dv/dt = 7. Use Analyze Statistics to find the mean value of acceleation. a 8. How well do you values of acceleation and initial velocity fom the diffeent gaphs match up? 9. Be sue to pint you gaphs when you have completed you analysis. Pat III Acceleated Motion Slowing Down. Investigating Integals. Pocedue 1. Disconnect the cat fom the pulley and emove the 500g weight. 2. Open the file 1DAcceleation_3.cmbl located in the Expeiments folde on the desktop. 3. Set the cat at the fa end of the senso s ange. Pactice giving it a single tap that is just stong enough to make the cat oll to a stop just in font of the senso. Note that the fiction between the cat and the table is causing the cat to negatively acceleate. 4. Mak the stating point of the cat. Have one goup membe hit the collect button. Then have anothe goup membe tap the cat. Measue the total distance the cat tavels using the ules. Repeat until you get a clean gaph. 4 of 15

5 Analysis Indefinite Integals 1. Use Analyze Statistics to find the mean value of acceleation. a 2. Wite out an equation fo acceleation as a function of time. This equation should be vey simple. a(t) = 3. Woking by hand, take the integal of both sides of the equation above. This should give you an equation fo velocity as a function of time. This equation should match the shape of you velocity vs. time gaph. v(t) = adt = 4. Use Analyze Linea Fit to fit a staight line to you velocity vs. time gaph. Detemine how each pat of the linea fit elates to the equation above, and fill in the numbes in the following table. a v 0 5. Take the integal of both sides of the equation above. This will give you an equation fo position as a function of time. This equation should match the shape of you position vs. time gaph. x(t) = vdt = 6. Use Analyze Cuve Fit to have the pogam find the paabola that best matches you position vs. time. data. Choose Quadatic, then click, Ty Fit. If the cuve seems to match you data well, click OK, othewise ask you instucto fo assistance. Identify how each pat of the cuve fit elates to the equation above, and fill in numbes in the table. a v 0 x 0 7. How well do you values of acceleation and initial velocity fom the diffeent gaphs match up? 5 of 15

6 8. Pint up you gaphs, then close the cuve fit and statistics boxes so that you page is clea fo the next section. Analysis Definite Integals 1. Highlight the good egion of data on you position vs. time gaph. Make sue that the exact same egion is also highlighted on the velocity and acceleation gaphs. You do not want to pay any attention to data that was collected befoe the cat was tapped o afte it came to a stop. 2. Click on the acceleation vs. time gaph and use the integate button on the toolba to pefom a definite integation. When you do this, the pogam is calculating the aea unde the cuve. Wite the numbe value of the integal below: adt = 3. Looking at the velocity vs. time gaph, use the examine button on the toolba to find the exact value of velocity at the endpoints of the highlighted egion. Calculate the diffeence between these two values: v = v f - v i = 4. Explain why the values fom pats 2 and 3 should be equal. How did you numbes match up? 5. Click on the velocity vs. time gaph and use the integate button on the toolba to pefom a definite integation. Wite the numbe value of the integal below: vdt = 6. Looking at the position vs. time gaph, use the examine button on the toolba to find the exact value of position at the endpoints of the highlighted egion. Calculate the diffeence between these two values: = x f - x i = This should be the total distance that you measued the cat to tavel using the ule. 7. Explain why the values fom pats 5 and 6 should be equal. How did you numbes match up? 8. Pint you gaphs when you have completed you analysis. 6 of 15

7 Conclusion Include an answe to the following question with you lab epot. A cat stats at est and olls down a amp as shown below. Gavity causes the cat to have a constant acceleation as it tavels along the amp. When it eaches the bottom of the amp, fiction with the flat suface causes it to gadually slow to a stop. Daw gaphs of the acceleation, velocity, and position of the cat thoughout its motion as a function of time. The gaphs should be laid out as they wee on the compute sceen, with the time axis fo all thee lining up. 7 of 15

8 Appendix: Theoy The basic vocabulay of kinematics o velocity, acceleation and time. consists of tems like position, displacement, speed Position is the easiest place to stat in the study of motion. The position vecto is simply the vecto that points to the place whee an object sits, in elation to some peviously defined coodinate system. Fo now, we will use a simple 1-dimensional Catesian coodinate Figue 1 The cente of this object sits (appoximately) at 2.5 units of dis tance in the positive x- diection. The position vecto, p, can be witten as p = 2.5 xˆ o p = 2. 5 i ˆ. It is customay to use the lettes iˆ, ˆ, j and kˆ to denote vectos in the x- y- and z-diections, espectively. When the object changes its position, we have Displacement. The displacement of an object is defined as the diffeence between its final and initial positions. Since the initial and final positions ae descibed by vectos, the displacement is also defined as a vecto. d = p f p i Eq. 1 This (pevious page) is a vey geneic statement. As you become moe familia with vectos and vecto notation, you may see statements like x = x x Eq. 2 whee x may epesent a vecto in 1 (î ), 2 ( i ˆ, ĵ ) o 3 ( iˆ, ˆ, j kˆ ) dimensions. Fo now we will wok in only one dimension. d x p f o f i p i p x x f x i x Figue 2 1 To see how position and displacement ae epesented in two dimensions, see Appendix A! 8 of 15

9 The small tiangle, the capital Geek lette delta, is a symbol epesenting change. The phase is ead The change in x is equal to x-final minus x-initial To find the value of the displacement vecto, we use vecto subtaction: d = p f pi d = ( 1ˆ) i (2.5ˆ) i d = ( 1 2.5)ˆ i d = 3.5ˆ i In tems of a vecto x we would wite: = x f = xi = ( 1ˆ) i (2.5ˆ) i = ( 1 2.5)ˆ i = 3.5ˆ i The numbe 3.5 tells us how long the displacement vecto is. The fact that it is negative tells us that the object moved in the diection of deceasing x. If we want to futhe discuss the motion of the object, we need to take into account the time it takes to displace the object. We will epesent the motion in 1-dimension by a set of coodinate axes, with the diection of motion along the vetical axis, and along the hoizontal axis we will put time. Thus we could have gaphs that look like this: a. x (m) b. x (m) The object is displaced by 1.5 metes in 2.5 seconds. The object is displaced by 1.5 metes in 2 seconds. c. x (m) d. x (m) The object is displaced by 1.5 metes in 1.5 seconds. The object is displaced by 1.5 metes in 1 second. 9 of 15

10 x (m) e. The object is displaced by 1.5 metes in 0.5 seconds. Figue 3 Displacement vs. Time gaphs Now we find ouselves in a position not only to discuss how fa an object was displaced, but also how long it took to move. Hee is whee we intoduce the concepts of Speed and Velocity. Speed and Velocity ae vey simila in the sense that they both tell us how fast an object moves how fa. The diffeence is that velocity is a vecto, and caies infomation about the diection of motion (fowad o backwad, East o Noth ) in addition to its magnitude. Speed is the magnitude of the Velocity vecto. Howeve, since we ae ight now woking with only one diection of motion, the tems speed and velocity ae nealy intechangeable. Velocity = the change in position of an object the time inteval in which it moved o v = t. Eq. 3 Thus, we can define velocities fo each of the gaphs pictued on the pevious page: a. v = 1.5m 2.5s = 0. 6m s b. v = 1.5m 2s = 0. 75m s c. v = 1.5m 1.5s = 1m s d. v = 1.5m 1s = 1. 5m s e. v = 1.5m 0.5s = 3m s 10 of 15

11 You should notice that the highe velocities coespond to moe steeply sloping distance vs. time gaphs, and the lowe velocities have moe shallowly sloping gaphs. This is not a coincidence! The definition of the slope of a gaph is m = y2 y1 x2 x1, whee y efes to the vetical axis and x efes to the hoizontal axis. This is the same as witing m = y. In ou case, the distance, x, is on the vetical axis, and time is on the hoizontal axis. This means that when we calculate the velocity, all we ae doing is finding the slope of the distance vs. time gaph. You may emembe fom calculus that the slope of a gaph is equivalent to the deivative of the vetical-vaiable with espect to the hoizontal-vaiable. This means that velocity is the deivative of distance with espect to time, o dx v =. Eq. 4 dt If you have a distance vs. time gaph that looks something like this: distance (m) v 1 v 2 v 3 v 4 v 5 Figue 4 Distance vs. Time Gaph you may take the slope (i.e. find the velocity) fo each of the staight-line potions of the gaph. If you have a situation like the one descibed above whee velocity is changing ove time, it may be useful to daw a velocity vs. time gaph. Fo the situation above, we would have something like this: v 2 velocity (m/s) v 1 v 3 v 4 v 5 11 of 15

12 Figue 5 Velocity vs. Time Gaph We can calculate how fa the object moves duing each velocity inteval by e-aanging the equation that defines velocity: v = t becomes = v t Eq. 5 If an object tavels with a velocity v 1 fom time t = 0 to a time t = t 1, the distance the object tavels is = v1 t = v1( t1 to ) = v1t1 If we visualize this calculation happening with the aid of a velocity vs. time gaph, we can see that the finding the poduct v 1t1 is equivalent to finding the aea bounded by a ectangle with sides of length v 1 and t 1. velocity (m/s) v 1 t 1 Figue 6 Calculating distance taveled with a Velocity vs. Time gaph Thus, the distance taveled by an object may be found by calculating the aea unde the velocity cuve (it is moe coect, paticulaly in the case of a negative velocity, to say the aea between the velocity cuve and the hoizontal axis. Howeve, the woding unde is most often used). velocity (m/s) v 1 t 1 Figue 7 12 of 15

13 You may emembe fom you calculus classes that the aea unde a gaph is equal to the integal of the vaiable on the vetical axis with espect to the vaiable on the hoizontal axis. Thus, distance taveled is the integal of velocity with espect to time. d = vdt. Eq. 6 This bings us to ou final definition of the day: Acceleation. Acceleation is any change in velocity. Since velocity is a vecto, and has components both of magnitude and diection, this means that eithe a change in magnitude o a change in diection of a velocity vecto (o both) is the esult of that object undegoing acceleation. Acceleation is quantified as the magnitude of the change in velocity divided by the time inteval in which the change occus: v a =. Eq. 7 t Acceleation, like velocity, is also a vecto and so has a diection as well as a magnitude. If the acceleation vecto points in the same diection as the velocity vecto, the magnitude of the velocity vecto will incease (the object will speed up). If the acceleation vecto points opposite (anti-paallel) to the velocity vecto, the magnitude of the velocity vecto will decease (the object will slow down). If the acceleation vecto points pependicula to the velocity vecto, the object will tun in a cicle without changing its speed. If the acceleation vecto points in any othe diection with espect to the velocity vecto, the object will undego some combination of change in speed and diection. As you may have guessed, acceleation, being defined by a change of velocity and a time inteval, may also be found gaphically as the slope of a velocity vs. time gaph. velocity (m/s) A Positive Acceleation velocity (m/s) A Negative Acceleation acceleation (m/s 2 ) acceleation (m/s 2 ) Figue 8 Velocity vs. Time gaphs and thei coesponding Acceleation vs. Time gaphs Systems with non-constant acceleation can be vey complex, and we will not be studying them in tems of kinematic equations of motion. Today, and fo the nea futue, we will be looking at systems with constant acceleation. cceleation (m/s 2 ) 13 of 15

14 Figue 9 Constant, positive Acceleation vs. Time The equation fo this acceleation is a( t) = a0. We simply have a constant acceleation that does not change in time. Since we defined acceleation as the deivative of velocity with espect to time, we can wite velocity as the integal of acceleation with espect to time: v( t) = a( t) dt = aodt Eq. 8 = a t + constant of integation We will define the constant of integation as v 0, the initial velocity of the object (which may o may not be zeo). Integating gaphically, we find that the change in velocity is equal to the aea unde the acceleation vs. time cuve. o acceleation (m/s 2 ) a( t) = a ( t) = v + a t 0 velocity (m/s ) v 0 v 0 0 v f v f v 0 = v Figue 10 Gaphical integation of Acceleation vs. Time yields change in velocity Integating the equation fo velocity will yield the equation fo the position: x( t) = ( v0 + a0t) dt = v0dt + a0dt 2 = v t + a t + constant of integation Eq of 15

15 In this case, the constant of integation is the initial x-position, x 0 zeo). (which again, may be Sim ilaly, gaphically integating velocity vs. time cuve gives the change in position: velocity (m/s ) v( t) = v0 + a0t distance (m/s ) x 0 x f 1 2 x( t) = x0 + v0t + 2 a0t Figue 11 Gaphical integation of Velocity vs. Time yields change in position x f x 0 = 15 of 15